0
$\begingroup$

I didn't find anywhere a simple function to describe a one-sheet hyperboloid based on arbitrary coordinates of the foci and the difference of distances between them.

So given the two points $F_1\, (x_1, y_1, z_1)$, $F_2\, (x_2, y_2, z_2)$ and the difference of distances $d \in \mathbb{R}$, I can define the hyperboloid as a locus of points $P(x,y,z)$ where $|PF_1| - |PF_2| = d$ that is

$\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2} - \sqrt{(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} = d$

The surface is closer to $F_1$ if $d<0$ and closer to $F_2$ if $d>0$. For $d=0$ it degenerates to a plane.

Is there a simpler form to describe this hyperboloid?

$\endgroup$
4
  • $\begingroup$ Technically, this looks to me like a hyperboloid of two sheets, even if you’re only interested in one of its components. It’s the surface of revolution of one branch of a hyperbola. $\endgroup$
    – amd
    Jan 31 '18 at 23:16
  • $\begingroup$ Thank you @amd, are you suggesting a parametrization? $\endgroup$
    – Guglie
    Jan 31 '18 at 23:19
  • $\begingroup$ I’d say construct a hyperbola in the $x$-$z$ plane with the foci on the $z$-axis, turn it into a surface of rotation by replacing $x^2$ with $x^2+y^2$ and then rotate and translate into position. I don’t see any pretty way to select only one branch of it without introducing square roots, though. That you could do more cleanly via parameterization. $\endgroup$
    – amd
    Jan 31 '18 at 23:23
  • $\begingroup$ Thanks @amd, I thought something similar, but I couldn't get it right, can you answer below so I can upvote you? $\endgroup$
    – Guglie
    Jan 31 '18 at 23:27
1
$\begingroup$

Your surface is one sheet of the surface generated by rotating a hyperbola about its transverse axis. The simplest representation is, I think, parametric. I would construct it in “standard position” first, and then rotate and translate it into place.

Let $f=|F_2-F_1|$. The hyperbola branch $x=\frac12\sqrt{f^2-d^2}\sinh t$, $z=\frac12 d\cosh t$ in the $x$-$z$ plane matches the given parameters, with the focus corresponding to $F_1$ on the negative $z$-axis. Rotating it about the $z$-axis turns this into $$x = \frac12\sqrt{f^2-d^2}\sinh t \cos\phi \\ y = \frac12\sqrt{f^2-d^2}\sinh t \sin\phi \\ z=\frac12 d\cosh t.$$ From here it’s a matter of rotating and translating it into place, which I’ll leave to you to work out. We really only care about the image of the $z$-axis, so taking advantage of the rotational symmetry it might be easier to construct a reflection that produces the desired tilt instead of rotating. This reflection would be in the angle bisector of the $z$-axis and $F_2-F_1$. (You might then want to replace $\phi$ with $-\phi$ in the parameterization to maintain orientation.) You can find some other parametric equations of a hyperbola to use as starting points here.

Alternatively, you could start with the implicit equation $${4z^2 \over d^2}-{4(x^2+y^2) \over f^2-d^2} = 1$$ of the above hyperbola and transform that, but I don’t see a good way to select one sheet of the surface without introducing square roots of the terms in the equation, which is what it appears you’re trying to avoid.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.